Let $g(x)=\dfrac{x^2-x-12}{x-4}$ when $x\neq 4$. $g$ is continuous for all real numbers. Find $g(4)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-3$ (Choice B) B $4$ (Choice C) C $-4$ (Choice D) D $7$
Explanation: $\dfrac{x^2-x-12}{x-4}$ is continuous for all real numbers other than $x=4$ which means $g$ is continuous for all real numbers other than $x=4$. In order for $g$ to also be continuous at $x=4$, the following equality must hold: $\lim_{x\to 4}g(x)=g(4)$ We will obtain the above equality by letting $g(4)=\lim_{x\to 4}g(x)$. So let's find $\lim_{x\to 4}g(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to 4}g(x) \\\\ &=\lim_{x\to 4}\dfrac{x^2-x-12}{x-4} \gray{\text{This is the rule for }x\neq 4} \\\\ &=\lim_{x\to 4}\dfrac{\cancel{(x-4)}(x+3)}{\cancel{(x-4)}} \gray{\text{Factor}} \\\\ &=\lim_{x\to 4}{(x+3)} \gray{\text{Cancel common factors}} \\\\ &\text{(This is allowed because }x\neq 4) \\\\ &=4+3 \gray{\text{Direct substitution}} \\\\ &=7 \end{aligned}$ We obtained that if we set $g(4)=7$, then $\lim_{x\to 4}g(x)=g(4)$, which makes $g$ continuous at $x=4$. Since we already saw that $g$ is continuous for any other real number, we can determine that it's continuous for all real numbers. In conclusion, $g(4)=7$.